If x + i y = (1 + i) (1 + 2 i) (1 + 3i), then x2 + y2 = A. 0 B. 1 C

1.	If x + i y = (1 + i) (1 + 2 i) (1 + 3i), then x2 + y2 = A. 0 B. 1 C. 100 D. None of these
Answer» Given that (1 + i) (1 + 2i) (1 + 3i) = x + i y …(1) We can also say that (1 - i) (1 - 2i) (1 - 3i) = x - i y …(2) Multiply and divide the eq no. 2 with eq no. 1 \(\frac{ (1 + i) (1 - i)(1 + 2i)(1 - 2i)(1 + 3i)(1 - 3i)}{(1 - i) (1 - 2i) (1 - 3i)}\) = \(\frac{(x + i y)(x - i y)}{x - i y}\) ((1)² – (i)²)((1)² – (2i)²)((1)² – (3i)²) = ((x)² – (iy)²) x² + y² = 2 × 5 × 10 = 100

If x + i y = (1 + i) (1 + 2 i) (1 + 3i), then x2 + y2 = A. 0 B. 1 C. 100 D. None of these

Answer»

Given that (1 + i) (1 + 2i) (1 + 3i) = x + i y …(1)

We can also say that

(1 - i) (1 - 2i) (1 - 3i) = x - i y …(2)

Multiply and divide the eq no. 2 with eq no. 1

\(\frac{ (1 + i) (1 - i)(1 + 2i)(1 - 2i)(1 + 3i)(1 - 3i)}{(1 - i) (1 - 2i) (1 - 3i)}\) = \(\frac{(x + i y)(x - i y)}{x - i y}\)

((1)² – (i)²)((1)² – (2i)²)((1)² – (3i)²) = ((x)² – (iy)²)

x² + y² = 2 × 5 × 10 = 100